my solution to jonsen's egg puzzle variant

eggs.py

def memo_cyclic(f_cycle):
    def decorator(f):
        cache = {}
        def f_memo(*args):
            args = tuple(args)
            if args not in cache:
                cache[args] = f_cycle(*args)
                cache[args] = f(*args)
            return cache[args]
        f_memo.__name__ = f.__name__
        return f_memo
    return decorator

fail = float("infinity")

@memo_cyclic(lambda *args: fail)
def f(n, lo, hi, fl=0, dr=0):
    if lo == hi:
        return 0 # only one floor is possible
    if n+dr == 0:
        return fail # out of eggs
    if fl == 0:
        n, dr = n+dr, 0 # pick up any dropped eggs
    return 1 + min(f(n, lo, hi, fl-1, dr) if fl > 0 else fail,  # go down one floor
                   f(n, lo, hi, fl+1, dr) if fl < hi else fail, # go up one floor
                   max(f(n-1, lo,   fl, fl, dr),                # drop egg (broken)
                       f(n-1, fl+1, hi, fl, dr+1))              # drop egg (unbroken)
                       if n > 0 and lo <= fl < hi else fail)

import sys
sys.setrecursionlimit(10000)

print [f(n, 0, n) for n in range(20)]
print [f(1, 0, n) for n in range(20)]
print f(2, 0, 99)